Uniform convergence of Fourier series

We study some of the basic properties of the Fourier series. In particular, we show that if the series of the Fourier coefficients of $f$ converges absolutely, then Fourier series $S[f]$ converges uniformly to $f$ at the continuous points of $f$.

Lemma (Condition for the uniform convergence of Fourier series)

Let $f\in {\mathcal{R}}_{2\pi }^1$ with its Fourier coefficients satisfying the conditions

$$\begin{array}{rcl}\sum _{n=0}^{\mathrm{\infty }}|a_n|& <& +\mathrm{\infty },\\ \sum _{n=1}^{\mathrm{\infty }}|b_n|& <& +\mathrm{\infty }.\end{array}$$

Then, the Fourier series $S[f]$ converges uniformly with respect to $x\in \mathbb{R}$. In particular, $S[f]$ is continuous.

Proof. Consider the following positive (non-negative) term series:

$$\frac{1}{2}{a}_0+\sum _{n=1}^{\mathrm{\infty }}(|a_n|+|b_n|).$$

By assumption, this series converges. Furthermore, this is a dominating series of $S[f]$. Thus, $S[f]$ converges uniformly. All terms of $S[f]$ are continuous (they are just $\cos$ and $\sin$). Thus, by the Theorem on Uniform convergence and continuity, $S[f]$ is continuous. ■

See also: Sequence of functions (Uniform convergence and continuity)

Lemma (Zero function)

If all the Fourier coefficients of $f\in {\mathcal{R}}_{2\pi }^1$ are zero, that is,

$$\begin{array}{}\text{(eq:zero)}& \{\begin{array}{ll}{\int }_{-\pi }^{\pi }f(x)\cos (nx)dx=0& (n=0,1,\cdots ),\\ {\int }_{-\pi }^{\pi }f(x)\sin (nx)dx=0& (n=1,2,\cdots ),\end{array}\end{array}$$

then $f(x_0)=0$ where $x_0$ is any continuous point of $f$. If, in addition to the condition (eq:zero), $f(x)$ is a continuous function, then $f(x)=0$ (i.e., identically zero).

Proof. By changing variables $x\mapsto x-x_0$, we may assume $x_0=0$. Thus, we may assume that $f(x)$ is continuous at $x=0$, and we show that $f(0)=0$. We prove this by contradiction.

Suppose $f(0)\ne 0$, in particular, $f(0)>0$. Then, Since $f$ is continuous at $x=0$, there exists a $\delta >0$ such that, if $|x|<\delta$, then $|f(x)-f(0)|<\frac{f(0)}{2}$. Combining with the triangle inequality $|f(0)|-|f(x)|\le |f(x)-f(0)|$, we have

$$f(x)\ge \frac{1}{2}f(0).$$

Let us define the following auxiliary function

$$p(x)=1+\cos x-\cos \delta .$$

Assuming $|x|\le \pi$,

$$\begin{array}{rcl}|x|\le \delta & ⟹& \cos x\ge \cos \delta ⟹p(x)\ge 1,\\ |x|>\delta & ⟹& -1\le \cos x<\cos \delta ⟹-\cos \delta \le p(x)<1\\ \text{(eq:px1)}& & ⟹& |p(x)|<1.\end{array}$$

For $k\in \mathbb{N}$, $p(x{)}^k$ is a $k$-th degree polynomial in $\cos x$. Meanwhile, ${\cos }^{k}x$ can be expressed as a linear combination of $1$, $\cos x$, $\cos (2x)$, $\cdots ,\cos (kx)$ (why?). Therefore, by (eq:zero),

$${\int }_{-\pi }^{\pi }f(x)p(x{)}^{k}dx=0(k=1,2,\cdots ).$$

Splitting the interval of the integral on the left-hand side into $|x|\le \delta$ and $|x|>\delta$,

$$0={\int }_{|x|\le \delta }\cdot +{\int }_{|x|>\delta }\cdot \ge {\int }_{|x|\le \delta }\frac{f(0)}{2}dx+{\int }_{|x|>\delta }f(x)p(x{)}^{k}dx.$$

Hence,

$$\begin{array}{}\text{(eq:0px)}& 0\ge f(0)\delta +{\int }_{|x|>\delta }f(x)p(x{)}^{k}dx.\end{array}$$

We now calculate the limit of the above integral as $k\to \mathrm{\infty }$. If $\delta <|x|<\pi$, by (eq:px1), we have

$$|f(x)p(x{)}^k|\le |f(x)|,$$

and

$$\begin{array}{rcl}{\int }_{|x|>\delta }|f(x)|dx& <& +\mathrm{\infty },\\ \underset{k\to \mathrm{\infty }}{lim}f(x)p(x{)}^k& =& f(x)\cdot 0=0(|x|>\delta ).\end{array}$$

By Arzelà's theorem,

$$\underset{k\to \mathrm{\infty }}{lim}{\int }_{|x|>\delta }f(x)p(x{)}^{k}dx=0.$$

Applying this to (eq:0px), we have $0\ge f(0)\delta >0$ which is a contradiction. Therefore, $f(0)>0$ is impossible. Similarly, we can show that $f(0)<0$ is impossible. Thus, $f(0)=0$. ■

See also: Arzelà's theorem

Theorem (If $S[f]$ is uniformly convergent, then $f=S[f]$)

If the Fourier series $S[f]$ of $f\in {\mathcal{R}}_{2\pi }^1$ converges uniformly, then $S[f](x_0)$ converges to $f(x_0)$ where $x_0$ is any point at which $f$ is continuous. In particular, if $f$ is continuous with a period $2\pi$ and $S[f]$ converges uniformly, then $f=S[f]$.

Proof. Let us define

$$g(x)=S[f](x)=\frac{1}{2}{a}_0+\sum _{n=1}^{\mathrm{\infty }}\{a_n\cos (nx)+b_n\sin (nx)\}.$$

Since the series on the right-hand side converges uniformly (by assumption), $g(x)$ is continuous, and term-wise integration is allowed for the functions $g(x)\cos (nx)$ and $g(x)\sin (nx)$. Noting the orthogonality of sine and cosine functions, we have

$$\begin{array}{rcl}{\int }_{-\pi }^{\pi }g(x)\cos (nx)dx& =& \pi a_n(n=0,1,\cdots ),\\ {\int }_{-\pi }^{\pi }g(x)\sin (nx)dx& =& \pi b_n(n=1,2,\cdots ).\end{array}$$

Combining these with the definition of Fourier coefficients of $f$, we have

$$\begin{array}{rcl}{\int }_{-\pi }^{\pi }\{f(x)-g(x)\}\cos (nx)dx& =& 0(n=0,1,\cdots ),\\ {\int }_{-\pi }^{\pi }\{f(x)-g(x)\}\sin (nx)dx& =& 0(n=1,2,\cdots ).\end{array}$$

Applying the above Lemma (zero function) to the function $f-g$, we have $f(x_0)-g(x_0)=0$ where $x_0$ is a continuous point of $f$. ■

The next theorem should now be trivial.

Theorem

Let $f\in {\mathcal{R}}_{2\pi }^1$. If the Fourier coefficients $a_n$ and $b_n$ of $f$ satisfy $\sum _n|a_n|<+\mathrm{\infty }$ and $\sum _n|b_n|<+\mathrm{\infty }$, then the Fourier series $S[f]$ of $f$ converges uniformly and $S[f](x)=f(x)$ at each continuous point $x$ of $f$.

Example. As seen above, the Fourier series in the above Example (evenfunc) converges uniformly. Therefore,

$$1-\frac{|x|}{\pi }=\frac{1}{2}+\frac{4}{{\pi }^2}(\frac{\cos x}{1^2}+\frac{\cos (3x)}{3^2}+\frac{\cos (5x)}{5^2}+\cdots )(-\pi <x<\pi ).$$

Note that we now have "$=$" rather than "$\sim$." □